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Did the cusp in the Boeing 377/C-97/KC-97’s fuselage structure cause fatigue problems?
The Boeing 377 Stratocruiser - a civilian adaptation of the C-97, a military transport which would also later be developed into the KC-97 - were derived from the B-29/B-50 family by grafting a large, bulbous passenger/cargo compartment onto the B-29/50 fuselage; this created a sharp cusp in the fuselage cross-section, as can be seen in this photo of a half-built Stratocruiser:

(Image source)
Such a sharp discontinuity in the fuselage profile would be expected to serve as a stress concentrator in a pressurised aircraft, like the 377/C-97/KC-97 series, and cause accelerated fatigue cracking in and around the cusp, shortening the lifespan of these fuselages; did these aircraft, in fact, experience fatigue problems emanating from the fuselage cusp, and, if so, how were they dealt with?

The floor in these aircraft takes part of the stress caused by the cabin pressure. It might not be as bad as it looks. But I am no expert in this stuff by any means :D

It is not the stress concentrating type of sharp corner. That is the natural shape that allows the parts to be in a nice smooth tension at lowest energy. If you made an elastic rubber balloon with a divider membrane creating two chambers and inflated it you would get a similar load distributing shape.

You can actually see the same effect happening with soap bubbles when they touch and form a flat film connecting them. You can actually still see this type of fuselage in use in the Boeing 717 (Douglas/McDonnell Douglas DC-9). A lot of other narrowbody airliners also have it, but the outer panels fair over the intersection between the two fuselages.


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Why does the DHC-6 have only one set of elevator cables?
Air Moorea Flight 1121 crashed because its up-elevator cable broke, allowing the elevator to blow to its faired position and thus be unavailable to counteract the pitch-down moment caused by the flaps being retracted;1 the DHC-6 elevator control system has two cables (up-elevator and down-elevator, distinguished by what part of the aft-quadrant bellcrank they attach to), and, if one breaks, the tension in both is released and the elevator floats.  No redundancy is provided in the form of a second or third set of elevator cables, unlike how most airliners have double- or even triple-redundant flight controls.  Why does the DHC-6 have only a single set of elevator control cables, rather than two or three?

1: No matter that retracting one’s flaps should not do that.

The Twin Otter isn't certified in a category that requires duplicate control runs with split surfaces.  It's more or less a single engine Otter with two PT-6s and a redesigned tail.  Those requirements apply to heavier aircraft than the DHC-6.

If it were a one-person general-aviation aircraft, I'd buy that explanation, but this is an airliner we're talking about here (albeit a small one).

The Beech 1900 is the Twin Otter's closest competitor in the commuter airline business, and it also has a single cable circuit.  You don't see separate left and right elevator control systems with pitch disconnects until you are in the Transport Category airplanes of the Dash 8/ATR size.

@Sean doesn't matter. The requirements are based on aircraft size, weight, etc. and not so much the purpose it's used for.

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Why aren't there any widebody propliners for the high-density short-haul markets?
Jetliners come in both narrowbody (one aisle) and widebody (two or more aisles) versions, but all propliners, to the best of my knowledge, are narrowbody aircraft.
Now, granted, one of the big reasons for having widebody airliners is to have an aircraft big enough to cross continents and/or oceans with a profitable passenger load, and trans[continental/oceanic] routes are (at least in civil aviation) exclusively the province of jets, because speed.
However, in places with lots and lots of huge, densely-populated cities packed fairly closely together, such as much of East Asia, passenger numbers are so high as to require the use of large widebodies even on very short-haul routes - and, on short-haul flights, propliners hold a considerable advantage over jets (jets are horribly inefficient in the dense air at low altitudes, and short-haul flights aren't long enough for an airplane to have time to climb to long-haul cruising altitudes, whereas propeller aircraft, with their lower speeds, and, thus, lower drag, do just fine going low, and short-haul flights are short enough that slower aircraft aren't a major disadvantage; this is why, even though propliners are essentially extinct in long-haul airliner service, they continue to take a large chunk of the short-haul market).  A widebody propliner would be perfectly-suited for these sorts of jam-packed short-haul markets.
So why aren't there any?

Why would it even be efficient to use airplanes for such short-haul routes?

Possible duplicate of Why are propeller engines uncommon on airliners?

@jamesqf: Well, at least one manufacturer thought it made enough sense to build and sell a short-range, high-capacity variant of one of its airliners specifically targeted at that market...

Noise and insufficient demand
Prop driven aircraft have much louder cabins than turbofan driven aircraft. Prop noise is directional, and the passenger cabin adjacent the blades is the most noisy position. Cockpit noise measurement is not representative.
The A400m has a fuselage diameter equivalent to the a330 and a gross weight similar to a 767, contrary to claims of a narrowbody cabin width measured to accomodate nine foot tall trucks. The AN22 has a maximum takeoff weight of 550klb, heavier than any a330. Both of these are currently flying turboprop widebody class aircraft; neither of these can carry passengers without hearing protection.
The A400m requires noise deadening treatment to even get to safe industrial levels of sound, let alone something that a passenger would be willing to endure. The AN22 is beyond ridiculously loud. Turboprop aircraft are aimed at short haul as passengers won't put up with the noise for more than a hour or two.
Any aircraft developed for long haul use needs to be quiet, and this is where widebodies aim. In the few situations where it is economical to use widebody aircraft for short haul routes (ie. Japan's 747D) a quiet widebody can be used with modest incremental cost. Conversely, a prop powered widebody could not be used on long haul with any conceivable current technology; passengers could not tolerate it.
Market demand does not justify the billions in development cost to create a short haul only widebody, especially given the ready availability of a substitute that passengers overwhelmingly prefer.

What makes you think that passengers wouldn't endure whatever noise level an aircraft produces?  Look at what they endure already...

Let us continue this discussion in chat.

Also, both the A400M and An-22 are military transport aircraft.  Militaries don't care if their soldiers have to wear earplugs for the entire flight, so they can afford to leave out the sound-blocking material and thereby gain a weight margin for carrying more troops.  If they'd been built for civil passenger service, you can bet your buttons they'd have sound-deadening material and a noise level just as good as any current jet.

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Why do so many UAVS use a V-tail configuration?
How different is V-tail from a conventional tail in the perspective of aerodynamics & flight dynamics of aircraft?
I have seen that many UAVs, like the General Atomics MQ1 Predator, LSI Amber, Mini Harpy, etc have a V-tail. Why is it so?
There are answers on how V-tail reduces the weight and drag but nothing on the dynamics of aircraft.

You say this as though reducing weight and drag don't impact the dynamics of an aircraft...

As you stated, V-tail has some advantages and some disadvantages when compared to conventional tail assembly. In case of UAVs the disadvantages are negligible, leading to them being a relatively common choice.
The advantages of V-tails are that they have fewer surfaces and (thus) fewer intersections, and they are lighter than a comparable conventional tail. All of these lead to less induced and parasitic drag, enhancing the efficiency of the overall design.
The downsides are susceptibility to yawing and more complex control arrangements with "fly-by-cable". To achieve the same stability as a comparable conventional arrangement, a longer rear fuselage is required.
On UAVs the control system is essentially direct drive, so control logic is managed by software instead of relatively complex arrangement of cables, pushrods and pulleys (as in the V-tail Bonanza [IIRC]). Software can also easily address any issue with stability, and since high maneuverability is not top priority in UAVs, control logic can be programmed as quite "conservative". Basically UAVs are not put into extreme flight conditions, so the adverse properties of V-tail are not an issue.

The mechanical control arrangement requires a "mixer", a little set of bellcranks to regulate pitch and yaw inputs/outputs, but beyond that though, the complexity isn't that significant.  The main flight behaviour issue is yaw excursions in turbulence because vertical gusts, that are offset off the vertical, impart a yaw force as one surface is affected more than the other.  This nose hunting in turbulence was the problem with the Bonanza.  With a FBW or yaw damping system, the issue is masked.

True, it's not all that complex @JohnK , but more complex than the conventional setup. I remember when I was a kid it was so inspiring trying to sketch out a cable and pulley setup that would execute proper movement of control surfaces

A friend was checking out his new glider purchase the other week, with a V tail, and it took a couple of minutes for him to figure out ruddervator movements when he was doing control checks, because he would move the surface while someone held the stick, but the rudder pedals would just give away.  He was pretty confused for a second or two lol.


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What is Neil Armstrong teaching in this image?
It seems to me that in this image Neil Armstrong is teaching something related with aerodynamics forces (lift, drag...), but I don't know exactly what is that "Triangle" that He drew (It doesn't seem to be an aircraft).

Once you get high enough and fast enough, don't worry about lift, just go forward fast and the ground just falls away from you.

Yes that is likely it.

He quite obviously is explaining his findings, that at certain altitude, if your angular speed=the rotational speed of the earth, then your speed equals your drag, and your weight is about half your altitude. This is known as the Armstrong-Stetson -paradox. –

@Jpe61 -- Can't wait to upvote your future answer.

He did fly the X-15, which flew fast enough to take into account the earths curvature.  Sort of half flying and half orbiting (which would reduce the lift requirement).  Angular speed = rotational speed of earth seems more applicable to geosynchronous orbit.  Would have enjoyed his classes.

I dare not @quietflyer

1)  The triangle shape represents an aircraft-- imagine something like a supersonic version of a Gee Bee racer.  The thing at the leading edge of the vertical fin is the canopy.
2) Here is a sketch that offers an educated guess about parts of the diagram that are either cropped, or hidden behind Neil Armstrong, in the original photos.  Also, for added clarity, labels have been added to the dashed lines representing the horizon, and the aircraft's longitudinal axis.  Note that for the diagram to make sense, the wings must be attached with zero incidence relative to the longitudinal axis of the vehicle, or the vehicle may be a flat-bottomed wingless lifting body.   Only the green vectors represent actual forces.  "L" is the lift vector, "D" is the drag vector, "W" is the weight vector, and "T" is the thrust vector.  Note that "D" is parallel to the flight path, "L" is perpendicular to the flight path, and "W" is perpendicular to the horizon. "T" is drawn in this sketch as nearly, but not quite, parallel to the aircraft's longitudinal axis-- in the original, the direction of the "T" vector is about midway between the direction of the aircraft's longitudinal axis and the direction of the flight path.  No attempt has been made in this quick sketch to make the force vectors add to zero.  We can't see enough of the original to guess whether the force vectors as drawn there would add up to approximately zero or not.
3) The vector labeled "flight path" in this sketch was labeled "V" (for "velocity") in the original.  It represents the aircraft's instantaneous direction (and speed) of travel through the airmass.
4) In the original, the line representing the orientation of the longitudinal axis of the aircraft was simply an extension of the flat bottom of the aircraft; it ended up getting displaced upward a bit in this sketch. This wasn't intentional but does have the effect of causing the line to pass closer to where the aircraft's CG would likely be. The angles of interest can be represented accurately regardless of whether or not the line representing the orientation of the aircraft's longitudinal axis actually passes through the CG or not-- we just need a reference line that is parallel to the aircraft's longitudinal axis.
5) Another detail that ended up different from the original in this quick sketch is the order of the angles illustrated on the left-hand side of the diagram.  In the original, running from right to left, the order is gamma, alpha (this letter has been rotated about 45 degrees clockwise so it looks almost like an upside-down gamma), theta, while the order in the attached sketch is theta, gamma, alpha.  This has no effect on the meaning of the diagram-- the same greek letters are still associated with the same illustrated angles. A confusing point about the upper photo is that it is cropped in such a way that at first glance the angle labeled "theta" appears to duplicate the angle labeled "gamma", but of course this is not actually the case.
6) It appears that the curved line represents the surface of the earth, the overlined "R" represents the radius of the earth, and "h" represents the aircraft's height above the earth's surface.  Neil is probably explaining that at very high speed, orbital effects start to come into play, reducing the need for lift.

Yes, it is rather clear (to me) that the curve is the surface of the Earth (and yes, h is height), and this is the crucial part of the whole thing. Neil is probably explaining that the Earth curvature matters for fast flight (and/or spaceflight). All other things are pretty standard and mundane.

It appears that he is teaching about the 4 fundamental forces of flight as well as the angle of attack, I also see notes about the thrust angle in the upper image
The triangle is just a crude airplane (I would think). Armstrong had time in a lot of airframes over the years including space craft, this just may be his generic scribble of a "flying machine", it could also just be an X-15 in the first image it appears there is some sketch of the surface of the planet.


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In an engine failure, should I expect less range than with the engine in idle?
I fly a C172 (fix pitch propeller) and when my instructor and I practice a simulated engine failure, we do not turn off the engine, instead we set the engine to idle.
In an actual engine failure, should I expect less range than with the engine in idle?

Related-- https://aviation.stackexchange.com/questions/64394/does-a-windmilling-propeller-create-more-drag-than-a-stopped-propeller-in-an-eng

You could make a "rate of descent" graph for various RPMs at your favorite approach speed while you were at it.  Especially if you rent, idle RPM may vary for each plane.  You should find a highest rate of descent somewhere where the prop is spinning, which could produce an "engine out/prop not spinning" rate of descent with a little power still in.  Keep in mind prop may still "wind mill" even if the motor shuts down.  This will produce the highest rate of descent (most net drag) at that airspeed (and weight).

Yes, but if your engine fails there will be many other factors that you would have to consider (available landing sites with safe approaches and acceptable surface, wind, etc.) that the difference in range should not be a major decision factor.

Yes to some degree.  It depends on the idle rpm, prop pitch, engine compression and your gliding speed, but if the engine is off but windmilling, there is substantial drag from the energy needed to windmill the prop without any help from the engine and it will knock some measurable amount off the glide ratio.
When it's idling, it might still be windmilling a bit if there is still a negative angle of attack on the blades.  At some rpm just above idle, the prop's rotation speed will be at the zero thrust point (blade AOA at zero) and the glide will be like there is no prop there at all. To know where that is, you'd have to be able to measure the thrust load directly somehow.  So it's quite a moving target, trying to come up with a gliding value with different conditions with a running engine.
With the engine off completely, the windmilling drag drag case is at its worst, and the best glide is achieved if the propeller can be stopped completely so that the blade is fully stalled with a near 90 deg AOA, but this is not something advocated in training.
A stationary fixed pitch propeller makes way less drag that one being driven around by airflow, which is effectively in an "autorotation" making substantial lift aft. You an visualize this by imagining an autogyro, which can descend vertically with the airflow going straight up through its rotor to drive it - it's windmilling - and the rotor creates enough lift to limit the gyro's descent rate to about 500 fpm. Make the rotor stop completely and the autogyro becomes a falling anvil.   Turn the gyro 90 degrees, make the rotor really small, like propeller size, and there's your gliding 172.
Problem is, while it's theoretically beneficial to get the prop to stop, you usually have to slow down quite a lot, maybe close to stall speed, not a good idea at low altitude. And a tired engine with low compression might keep windmilling right down to the stall speed (and an engine with tight cylinders may stop windmilling right at glide speed - a nice bonus).  Under a few thousand feet, there are too many risks and other priorities to bother with this just for the L/D improvement.
If the engine quit at a substantial altitude and glide distance was absolutely critical (like over the water and trying to make it back to shore), I might try slowing it down to get the prop to stop windmilling. How slow I have to go to get it to stop, and how fast I can go before it starts windmilling again depends on the cylinder compression in the engine, so it may work well or it may not, and I may have used up most of the benefit trying to get the thing to stop.
As a student pilot, don't go there on your own (thinking about slowing down to stop the prop) but it wouldn't hurt at all to ask for a demonstration at altitude to observe the glide performance with the prop windmilling vs stopped vs idling.  You'd get some sense of the drag differences by noting the descent rate at the same airspeed for each case.
The above applies to fixed pitch propellers.  If you have a constant speed prop (variable pitch, non feathering, and if it's feathering, like on a turboprop, well it's not an issue at all), it gets more nuanced.  Normally if the engine quits the prop goes to full fine pitch with the blades nearly flat and it likely will stop windmilling at gliding speed.  Some people argue that moving the prop control to full coarse (low RPM) before the oil pressure decays can get the blades to kind of "part way to feather" and this will reduce the drag of the prop to less than a stationary prop at full fine with the blades nearly flat.  On the other hand, the coarse pitch might just get it windmilling... Hard to say without some testing.

You're right.  I made some changes and added a para at the end, although I don't think we need to cover feathering single engine props on turbines, but I hope the clarifications help.  Thanks.

Much better.  Deleted my comment and upvoted.

Either the prop will windmill or it won't.  If it is windmilling, then it's spinning the engine.  If it windmills above programmed idle speed, the engine computer's idle-control system should turn off fuel injection altogether, since any use of fuel is a) unneeded and b) would increase speed even further beyond idle, and that's not the idle system's job.   Your car does that approaching every traffic light. Well my car does and it's 25yo.

Well first of all the poster's C-172 will have a Marvel Schebler carburetor, so if there is fuel and spark it's running and all the airflow will do is unload/backdrive the prop and raise the idle speed.  And if it is injected, aircraft fuel injection systems are ancient technology mechanical continuous flow systems excepting the very latest FADEC piston engine systems you might see on some high end stuff.  The regular stuff doesn't work that way.  Airplanes still use magnetos that are similar to tractor mags from the 30s.  In any case, we are mostly concerned with windmilling while failed.

@Harper-ReinstateMonica Most GA aircraft don't have FADECs. Or any sort of ECU. I'm 34 and I've literally never flown an airplane younger than I am. The newest one I've flown was built in 1980, but based on a design from the 60s. My experience is not particularly atypical in that regard for GA pilots, as far as I know.

The answer to this is a definite "maybe".
YouTube personality Trent Palmer tested this very question. He was actually able to glide further with the engine off than he was with it at idle. It was a very small gain, though, on the order of a few seconds extra flying time per 1,000 feet descent. The people he was with reported different results: One guy said he glided further with the engine at idle than with it off, but, again, the difference was miniscule. So, the ultimate answer is that it depends on your exact aircraft.
Edit: My answer originally said that everybody involved in the test got the same results. I was working from memory when typing this. Having rewatched the video, I see I was wrong in that, some people got different results (which actually illustrates the point better in my opinion).

I would bet that on any relatively draggy airframe like a Kitfox, or 172 for that matter, that the difference is maybe half a point in L/D if that.  Probably a lot more of a difference on a fast composite a/c  If you have a constant speed prop, and you think fast, it can be beneficial to pull the prop to MIN rpm while there is still residual oil pressure, and hopefully drive the blades to full coarse, getting you at least part way to feathered.  On the other hand, the coarse blades may just windmill easier, making things worse.  More testing required...

I feel that in most cases a partially powered prop will have a longer glide trajectory than a stationary or windmilling prop. Considering that when the prop is stationary, it effectively increases the frontal area of the aircraft, by the silhouette of the blades, but when windmilling, it in effect becomes a complete disk of drag, extracting energy that would otherwise be used to extend the glide.
In the case of a real engine failure, in a fixed pitch installation, windmilling or not becomes a case by case situation, depending on the nature of the failure. Further, there is a certainty that at sometime, the aircraft will get to the ground, with no chance of a go-around.
In a simulated engine failure, it is quicker, easier and more reliable to power up a warm idling engine than to initiate a restart of a potentially now very cold engine. Add to this, the associated costs of thermal shock resulting in cracked cylinder heads, is why simulated engine failures are simulated.


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How (and why) does engine thrust change with airspeed?
I'm interested in how the thrust of a turbofan engine is affected at higher airspeeds (TAS). I know (I believed) that engine thrust(at constant N1) was relatively constant like in the following graph (only slight deviations):

This graph is usually in the books/manuals describing engine performance with reference to speed.
Then I came across the data from CFM56-5C turbofan engine which states that max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N. That's almost 5 times more power on the ground than in cruise. Here is the link: How much air, by mass, enters an average CFM56 turbofan engine cruising per minute?
These are apparently contradictory statements or I am missing something. Which one is corrrect and why? Why is engine thrust being changed with speed? Also, on the graph above what are those two curves that when added form a net engine thrust?
After I did a few calculations using the thrust equation (F= mass flow * difference in exhaust and inlet velocities denoted as delta V -> we will disregard the fuel mass flow and assume exit pressure is equal to the free stream pressure thanks to a nozzle) and following data mentioned above in the link, I found out that delta V term in cruise and on takeoff is constant (at full power) and its value is 295 m/s, which states that exhaust velocity of the engine will always be 295 m/s faster from the inlet velocity(for a maximum power setting at any speed). I think that's logical because work done by the engine is used to increase kinetic energy (delta Ek) of the airflow which increases speed always by a constant amount at specific power/N1 setting (of course less power equals less delta V).

Please note the following: you state that max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N. That's almost 5 times more power on the ground than in cruise...  That's wrong, because thrust is not power. There exist a relation between thrust, speed, and power, but the stationary case is an special one...

In terms of useful work, $energy = force \cdot distance$,  $power = force \cdot velocity$ (vector dot products).  At zero speed, using power to blow all that air behind you costs a lot of fuel energy but isn't doing anything useful.  Even when you release the brakes and start to roll, it's not much power.  (Jet engines are very inefficient at low speed.)

The first diagram you link to shows three lines but does not indicate what they represent. I guess the bold line is thrust over speed. Then this diagram is correct for a turbojet.

Thrust $T$ is the difference between the engine's exit impulse minus the entry impulse: $$T = (\dot{m}_{air} + \dot{m}_{fuel})\cdot v_{exit} - \dot{m}_{air}\cdot v_{entry}$$
The exit speed $v_{exit}$ of a turbojet engine is almost constant over flight speed (relative to the engine of course), so as the engine accelerates, a larger entry impulse must be subtracted from a nearly constant exit impulse. Thrust drops slightly over speed.
At higher Mach numbers, precompression from the ram effect at the intake raises the pressure level (and hence the mass flow $\dot{m}_{air}$) inside the engine, so it will develop more thrust than in static conditions. This effect causes the thrust line to bend upwards at higher speed, and since the precompression grows nonlinearly with speed, the initial drop in thrust is soon reversed. Of course, now the fuel mass flow $\dot{m}_{fuel}$ will grow in the same way, so the fuel efficiency (thrust per fuel used) will continue to drop as speed increases.
Only when flight speed approaches the exit speed of the jet will thrust go down again. The typical exit speed of a turbojet is easily supersonic, therefore this type of engine is well suited for supersonic flight.

max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N

Here you have two effects combining to lower thrust. One is the reduction in the difference between entry and exit speed. This is more pronounced in a turbofan engine because the bypass flow will be accelerated much less than the core flow, and a higher flight speed will cause a proportionally bigger drop in thrust.
The second effect comes from the difference in air density between ground and cruise: Air density at a typical cruise altitude of 35,000 ft is only 0.38 kg/m³ or 31% of the air density at sea level. The original source for the cruise thrust number does not say for which altitude the figure is valid, but you can be sure that it is for about one third of ground density. Mass flow $\dot{m}_{air}$ is directly proportional with ambient density, and both effects combine. However, most sources give only a drop to a quarter of static thrust - the last table in the linked answer looks like someone mixed the values for the CFM56-5A and the CFM56-5C.

Just to clarify, by a turbojet's exit speed being "supersonic" is that meaning within the exhaust (with a higher speed of sound because it's hot) the actual flow is supersonic, or is it subsonic but faster than the speed of sound in the surrounding air?

@Talisker: Yes, supersonic means more than the speed of sound of this hot air. There are supersonic aircraft with subsonic engine exit speed, and that only works because of the higher speed of sound in hot air, but those are the exception.

@PeterKämpf when you said $v_{exit}$ is almost constant, is it constant relative to the aircraft so when the aircraft accelerates the exhaust velocity actually decreases relative to someone standing at the ground? or do you mean that the exhaust velocity is constant regardless of how fast the aircraft is going and so it actually increases relative to aircraft?

@AbanobEbrahim: Yes, very good question! The exit speed is approx. constant relative to the aircraft, so this puts an upper limit on the maximum speed which can be achieved.

@PeterKämpf: Thank you. I'm new here on Aviation and I already admire your awesome answers. But just to make sure I understand you correctly, let's assume the exhaust velocity when the engine is stationary is 500 $m/s$. Now by approx. constant you mean that when the aircraft is moving at 200 $m/s$, the exhaust speed relative to the aircraft is still 500 $m/s$ but to someone standing on the ground the exhaust speed is just 300 $m/s$ in the opposite direction of the aircraft. So is that correct?

@AbanobEbrahim: Yes, that is correct. For the nitpickers: The actual numbers might vary a bit between the different scenarios.

Air density or pressure altitude must be factored in when comparing engine thrust figures.  At cruise, FL400, air density will be about one fifth.  Max thrust (or max HP for piston engines) will also be one fifth, as will air resistance/parasite drag.  Air mass flow will be one fifth, so will fuel flow/burn.  Even lift will be one fifth for a given air speed (TAS).
I assume you meant TAS in your question, however, if we consider IAS/CAS, that is a different story.  At FL400, you would have numbers like Mach 0.82, 470kts TAS and 250kts IAS.  Airliners rarely do level flight except at cruise altitude, however it would be possible to maintain 250kts IAS/CAS with 20% of static thrust (give or take) at altitudes lower than cruise, such as in a hold.
Playing with these might be helpful:
IAS/TAS vs Altitude:  https://aerotoolbox.net/airspeed-conversions/
Engine power vs Altitude:  http://www.csgnetwork.com/relhumhpcalc.html

I'm not sure that this answers the question that was asked.  It's not bad information, but it's about how thrust varies with altitude, rather than what the question asks, which is why thrust varies with airspeed, per se.  Also, I'd question your assertion that an airliner could ever maintain speed with only 20% of available power.  Maybe on an incredibly light 747 or A-380, but on the 737's that I'm familiar with, that statement seems unsupportable.

Ralph J, you are misunderstanding me if you say "20% of available power". I am addressing the question which references one fifth of static thrust or 29,360N. By definition of what IAS is and what it implies, how can anyone take issue with my statement? If thrust is 29,360N in cruise at a particular indicated airspeed, both will be approximately the same for any other altitude (except of course for speeds and altitudes out of limits).

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What is this ring-shaped device behind this DC-8's jet engine?
What is this ring-shaped device behind this Douglas DC-8's jet engine?
(YouTube)

It's an ejector and was fitted on the DC-8's equipped with JT3C, JT4A or RR Conway engines. Normally extended for T/O and landing, it is retracted in flight after T/O because it is no longer needed and creates drag at higher speeds, and then it is extended during the descent/approach phase. It is operated hydraulically to extend and retract, and can be extended by an air bottle in the event of a hydraulic failure. The controls for extending/retracting are two switches on the left overhead panel, one for ejectors on the two outboard engines(1 and 4) and the other for the two inboard engines (2 and 3), and the backup compressed air bottle extends all four and is located on the left side of the throttle quadrant. Individually, the ejectors can be extended by pulling up individual thrust reverser levers on the throttles and the ejectors will retract again when reverse thrust is deselected. The ejector does two things. It reduces noise and improves thrust output up to approx. 150 knots. It also carries the thrust reverser which operates pneumatically to increase descent rate or slow the aircraft inflight and on landing to reduce the landing roll and reduce wear on the wheel brakes. It was developed and fitted on the DC-8 at the factory by Douglas during the flight test phase of the DC-8 program before certification and entry into service and was not an aftermarket hush kit attempt by any third party. I know this because I was a Captain on the DC-8 for 10+ years.

People got this one partly right; it's Xenu's ride - a DC-8 (not a 707) fitted with aftermarket hush kits to reduce noise.  They included a non-flush clamshell thrust reversers for the hot section fitted to accommodate the hush kit nozzle.

do we really need the scientology reference?

The early DC-8 fitted with Rolls-Royce Conway engines had that airframe mounted ejector type thrust reverser system, it was simple, fairly light and reliable.
Even though the ejector brought some noise attenuation benefits it was not part of a hush kit. The Conway engine was very responsive but also very noisy.
Later DC-8, fitted with Pratt & Whitney JT3D engines had engine mounted fan (bypass flow) reverser doors and engine mounted primary flow clamshell type reverser. These 2 systems were heavy, complex and hardly reliable.
Boeing 707 fitted with P&W engine also had complex, awkward thrust reverser systems.
Trans Canada Airlines was the ancestor of Air Canada and purchased 11 of the Conway powered DC8. It was the first engine I worked as a 19 year old apprentice mechanic at Air Canada power plant shop in Dorval in 1967.

Welcome to Av.SE!

They appear to be thrust reversers, similar to what you can see described in this book, and also similar to what is shown in this video of an early Boeing 737.
From what I've seen discussed in various forums the Boeing 707 had two thrust reversers; one for the cold section of the engine (example) and one for the hot section. As far as I can tell the second one was only used on a few early aircraft, as the control mechanisms required quite a lot of maintenance due to the high temperatures from the exhaust, but haven't found good sources to confirm this.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why do some aircraft (e.g. Avro RJ85) have rear-mounted air brakes?
I recently saw an Avro RJ100 on approach and I wondered why the Avro RJ100 and also the RJ85 have the air brakes at the rear of the aircraft, and not on top of the wings like most airliners?
(photobucket.com)
What is the advantage you get from this, and/or what made it impossible to mount the air brakes on top of the main wings?

They have them mounted at the rear fuselage because they work better, and because there isn't a set of stabilisers mounted there: the stabilisers are mounted on top of the vertical tail, in a T-tail arrangement.
The Fokker 100 has the same arrangement with air brakes as the RJ100 has. My office used to be at the airport where they were built, and I could see the line-up of landing aircraft coming in, all nose high - except for the little Fokkers which had their nose pointed at the runway allowing the pilots a clear vision unimpeded by a nose cone sticking up.
Wing spoilers acting as air brakes brake very efficiently as well of course, but they also destroy lift. In order to retain the amount of lift required to fly, the aircraft angle of attack needs to increase. That can be unsafe in a landing situation, so the manufacturer needs to mount leading edge slats which allow for higher AoA.
Air brakes such as the RJ100 has have only upsides, since they influence drag only, not lift. They're not used everywhere because the bit of fuselage where they are mounted usually houses horizontal stabilisers: deploying the air brakes would distort the air flow around these, which is never a good thing.

The Avro RJ100 does have something on its wings- three spoilers on each, though these act exclusively as lift dumpers after touchdown (and another two for roll control).
It is perfectly possible to make the lift spoilers act like airbrakes- as in other aircraft; in case of 146, having exclusive airbrakes serve a number of purposes:

The airbrakes increase the descent rate of the aircraft, more than doubling it compared to other airliners. This allows for steeper descents (like in London city).
The large area of the airbrakes also eliminates the need for complex thrust reversers once on the ground; this reduces maintenance requirements and cost.
The airbrakes also help in speed control. The use of airbrakes enables appraoch at higher thrust settings, enabling a better option of go-around rather than relying only on the throttle controls. Unlike the spoiler, the airbrake doesn't affect lift, enabling its use in such scenarios.
The airbrake is effective in all operational speeds, enabling faster descent in case of an emergency, though I'm not sure if it was a design consideration.


text:
Depth and range for FDR/CVR black box detection
What is the maximum depth range of a FDR underwater, at which it is able to be detected by receivers?
I want to know the maximum range of ULB pinger locator if it's submerged.

The maximum range if it's transmitting on 37.5kHz is usually between 1 and 2 kilometers in poor conditions and around 4 to 5 kilometers in normal conditions and up to 6 or 7 in good conditions. The range depends on the frequency it's transmitting but the ULB's work on 37.5kHz.


text:
What is "airspace underlying a Class B airspace area designated for an airport"?
For the purposes of airspeed restrictions within airspace, the FAA regulations list the following area description (emphasis mine):
14 CFR 91.117 (C)

No person may operate an aircraft in the airspace underlying a Class B airspace area designated for an airport [...] at an indicated airspeed of more than 200 knots (230 mph).

Should we understand the described airspace to be any and all airspace underlying a Class B where that specific Class B is designated for an airport? Or, should we understand the described airspace to be any and all airspace designated for an airport (Class C, D, or E) which is underlying a Class B?
Are there letters of interpretation or other regulations which clarify the grammatical ambiguity of this regulation?

Yes, there is a legal interpretation on this:

[...] under  the  express  language  of  the  regulation,  aircraft
  operating  in  the  airspace  underlying  Class  B  airspace -
  irrespective  of  whether  the  underlying  airspace  is  controlled
  or  uncontrolled - may  not  exceed  the  200-knot  speed  limitation

So it is indeed any and all airspace underlying class B.
And note the exception to the rule:

We  note  §
  91.117  (d)  does  provide  that,  if  the  minimum  safe  airspeed  for  any  particular  operation  is  greater  than  the  maximum
  speed  prescribed  in  §
  91.117,  then  an  aircraft  may  be  operated  at  that  minimum  speed.

Therefore, the retention of the phrase, "designated for an airport" is superfluous and is to blame for the ambiguity of the regulation.

Indeed.  "Designated for an airport" does appear in some of the descriptive language in the paragraph introducing class B airspace in FAA Order 7400.11C, but there was no reason to put it into that FAR.  Likewise with some other regs-- see for example https://aviation.stackexchange.com/questions/65697/a-question-about-far-91-225e2-pertains-to-ads-b-out-requirement-and-aircr .  However there are some other regs where the inclusion of the phrase "designated for an airport" arguably functions specifically to exclude E4 (Class-E-to-surface "extension") airspace and thus is not superfluous.

If we were including parenthesis is in algebra equations, the phrase would like like this:  (the airspace underlying (a Class B airspace area designated for an airport))   .  "Designated for an airport" modifies "a Class B airspace area" rather than modifying "the airspace underlying".  And yes it appears to be completely superfluous here.

To add to Pondlife's answer, the reason for this rule is that many VFR aircraft will try to fly under Class B airspace to avoid dealing with ATC so the airspace underneath the shelves of class B can be congested.  Therefore the FAA put this lower speed limit in place to allow for better VFR separation of aircraft under Class B airspace.


text:
What is the optimal bank angle to accomplish a given turn in a glide?
For an aircraft in a glide, what—if any—is the optimal bank angle to make a given turn while maximizing glide distance and maintaining published best glide speed?
In other words, at a constant airspeed, which bank angle will produce the greatest turn rate with the smallest loss in altitude. Assume a constant airspeed turn that is already established. I am not asking which bank angle will produce the least reduction in the vertical component of lift.
In envisioning this scenario, I specifically have in mind a single engine aircraft experiencing an engine failure at low altitude. However, the aerodynamic theory may apply to any fixed wing scenario.
Edit: let me emphasize that this question assumes the pilot is maintaining the aircraft's nominal published best glide speed throughout the turn, not adjusting airspeed to some variable theoretical best glide speed. Airspeed should not be a variable.

this is a very tricky question because so many factors come into play.

@rbp There are certainly a lot of factors that go into describing the scenario, but I think the solution is probably rather simple. One of my instructors, who was a former Boeing engineer, used to tell me that the answer was 45°. If that is true, I'm looking for the why.

I don't think the answer is aircraft dependant. It has to do with the trig of the lift vector and the rate of turn vice altitude loss.

@rbd Valid point in and of itself, but I don't think glide ratio intersects with the issues at play in this question.

Given the last paragraph, this might be duplicate of http://aviation.stackexchange.com/q/786/524, since that question does discuss the bank angle and speed to use for the turn.

Why are people so eager to close questions on SE Aviation. e.g. This is an excellent question with no obvious answer as far as I can see anywhere on SE Aviation nor on the internet. Yet, some trigger happy fellow wants to close it already as a duplicate.

Obviously the OP is free to constrain the question however he wishes.  But once you decide you wish to be travelling in a different direction than you are presently going, maximizing the distance travelled per altitude lost is not necessarily a significant consideration.  Certainly not if you are going to turn a full 180 degrees.  In that case, min sink speed might be more appropriate during the turn.  Yet min sink speed, just as max L/D speed, is a function of bank angle.  So staying at the wings-level max L/D speed might not be such a bad idea--

Please clarify the specific goal. All turns have radius in addition to turn rate, a 180 degree turn does not put you back where you started facing the other direction. A 90 degree turn can be started sooner or later to match the radius with desired exit track. Just eyeballing it you would want a fairly steep initial turn followed by a slow rollout kind of like a chandelle, this would quickly arrest movement in the original direction while sacrificing some glide ratio, then restore the glide ratio as soon as you are moving in the desired general direction. A real answer is knee-deep in calculus

It seems to me that it depends on the degrees of turn you want to accomplish.  If it's 1 degree, the answer is obviously essentially the best L/D speed.  If it's 180 degrees, then it's obviously not.  In other words, is motion in your original direction still helpful to some degree?  Or is it taking you further from a safe landing field?  Doesn't that have to affect the answer?

The best bank angle is indeed 45°.
It can be shown relatively easily that that gives the tightest turn for constant angle of attack, where we assume that constant angle of attack results in constant angle of descent too. It would be pretty hard to show for constant speed, and it might not even be exactly true in that case.
However, this can't really be separated from the best speed. And the surprising answer is that the stall speed (which is ~19% ($\sqrt[4]{2}$) higher than in straight flight) gives smallest loss of altitude for turning given number of degrees.
The reason is that the drag (near the stall speed) is proportional to $\frac{1}{v}$, but the turn radius is proportional to $v^2$, so as you slow down, the radius decreases faster then the drag (and therefore vertical speed) increases.
References:

http://www.nar-associates.com/technical-flying/impossible/possible.html
Is it even remotely feasible to turnback a single engine aircraft with an engine failure?

"on the back of the power curve"? I thought this was an engine out scenario

@rbp, "back of the power curve" is simply below best glide speed, i.e. in the range where lower speed means higher drag and whether the power to overcome that drag comes from engine or from descending is secondary.

i'll tell that to my glider pilot friends. I think you mean "back of the lift curve?"

@rbp, somehow "lift curve" makes less sense, because the plot is drag vs. speed, where drag can be expressed as power (for powered planes) or as vertical speed (for gliders; because vertical speed (times gravity) is power).

@rbp The term "power curve" refers to the fact that the curve is a polynomial -- i.e. Y equals X raised to a power (or negative X, times X raised to a different power, plus a constant, etc.etc.etc.). It has nothing to do with engine power.

@hiddenwindshield wrong on several different levels :) when used to describe a generic function, "power" usually refers to an exponential, not a polynomial. The curve of D/L as a function of airspeed is neither an exponential not a polynomial, and its name does come from "power required for lift to equal weight". That power can come from an engine or from expending gravitational potential energy or any combination.

The question as asked is open to interpretation, so I will first rephrase it to have a basis to build upon. Your last paragraph tells me that you want to know the optimum bank angle to get the highest ratio of turn rate to altitude loss in a glide at a given airspeed.
Spoiler: Since steeper bank angles require more lift, and aircraft with better L/D are more efficient in producing lift, the optimum bank angle depends on the aerodynamic qualities of the aircraft.
What is given

Glider or powered aircraft with inoperative engine. The polar and the weight are known and do not change over time.
Airspeed. This will result in a restricted optimum - the absolute best bank angle will require a suitable speed.

What can be changed

Bank angle $\varphi$ (obviously - you are asking for this)
Lift $L$ (again, obviously. You want to stay airborne)

Solution
First I need to formulate the ratio of turn rate over height loss. This then needs to be derived with respect to the bank angle and set to zero. To have a derivable polar, I use the quadratic polar where $c_D = c_{D0}\cdot\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$.
I further assume a coordinated turn, so we can define the lift and drag equations. Drag is compensated by selecting a suitable glide path angle $\gamma$ in order to convert potential into kinetic energy to keep the speed constant. The angular velocity $\Omega$ in a turn with the radius $R$ is
$$\Omega = \frac{v}{R} = \frac{g\cdot tan\varphi}{v} = \frac{g\cdot \sqrt{n_z^2-1}}{v}$$
Height loss over time is vertical speed $v_z$, and this can be calculated from speed $v$ and flight path angle $\gamma$:
$$v_z = v\cdot sin\gamma$$
Since $v$ is given and constant, we can rephrase the problem as a maximization of turn rate over flight path angle or sink speed. This is equivalent to the smallest height loss for a given azimuth change.
$$\frac{\Omega}{v_z} = \frac{g\cdot tan\varphi}{sin\gamma}$$
Before deriving this, we need to express $\gamma$ in terms of $\varphi$. If we had the liberty to adjust speed, we could directly solve for the optimum bank angle at optimum L/D. Now, however, speed is fixed and L/D is what the airplane produces at the required lift. Since for gliders $sin\gamma = \frac{c_D}{c_L}$, we can write:
$$\frac{\Omega}{v_z} = \frac{g\cdot tan\varphi\cdot c_L}{c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}} = \frac{g\cdot sin\varphi\cdot \frac{m\cdot g}{q\cdot S}}{c_{D0}\cdot cos^2\varphi + \frac{\left(\frac{m\cdot g}{q\cdot S}\right)^2}{\pi\cdot AR\cdot\epsilon}}$$
with $c_L = \frac{m\cdot g}{q\cdot S\cdot cos\varphi}$. Since the dynamic pressure $q$ is constant, we can now derive with respect to the bank angle. With the chain rule we get a fraction, and since it will be set to zero, it is enough to look for the condition when the numerator is zero:
$$g\cdot cos\varphi\cdot \frac{m\cdot g}{q\cdot S}\cdot\left({c_{D0}\cdot cos^2\varphi + \frac{\left(\frac{m\cdot g}{q\cdot S}\right)^2}{\pi\cdot AR\cdot\epsilon}}\right) = g\cdot sin\varphi\cdot \frac{m\cdot g}{q\cdot S}\cdot 2\cdot c_{D0}\cdot sin\varphi\cdot cos\varphi$$
$$\frac{\left(\frac{m\cdot g}{q\cdot S}\right)^2}{c_{D0}\cdot\pi\cdot AR\cdot\epsilon} = 2\cdot sin^2\varphi - cos^2\varphi = \frac{1}{2} - \frac{3}{2}cos2\varphi$$
$$\varphi = \frac{1}{2}\cdot arccos\left(\frac{1}{3} - \frac{2\cdot\left(\frac{m\cdot g}{q\cdot S}\right)^2}{3\cdot c_{D0}\cdot\pi\cdot AR\cdot\epsilon}\right)$$
This does not obviously look wrong, but I could very well have screwed up on the path to the result. If you plug in the numbers for an airplane you know, you can check whether the result makes sense. At least, with too low airspeed you get a negative argument for the cosine which mathematically means a roll angle of >90° and can be interpreted as too slow for that turn.

EDIT
Now we got a similar question but with both speed and roll angle as variables. Obviously, now we need to derive both with respect to speed and roll angle. But it is more fun to plot the results over these two as a contour plot. I just had to do this since several answers here claim that the optimum angle is 45°. Equally obviously, this is too simplistic.
First the math: I start from the same equations as above and added a term for wind ($w_z$) which adds rising or sinking air mass to the problem.
$$∆h = \frac{\pi\cdot v}{g\cdot tan\phi}\cdot(v_z+w_z) = \frac{\pi\cdot v}{g\cdot\sqrt{n_z^2-1}}\cdot\left(\frac{v\cdot c_D}{c_L}+w_z\right)$$
Expressing the lift coefficient as $$c_L = \frac{2\cdot n_z\cdot m\cdot g}{\rho\cdot S\cdot v^2}$$
brings us to $$∆h = \frac{\pi}{g^2\cdot\sqrt{n_z^2-1}}\cdot \left(\frac{\rho\cdot S\cdot v^4\cdot c_{D0}}{2\cdot n_z\cdot m} + \frac{2\cdot n_z\cdot m\cdot g^2}{\pi\cdot\rho\cdot S\cdot AR\cdot\epsilon} + w_z\cdot v\cdot g\right)$$
Nomenclature:
$\kern4mm g\kern6mm$gravitational acceleration
$\kern4mm n_z\kern4mm$vertical load factor
$\kern4mm \rho\kern6mm$air density
$\kern4mm S\kern5mm$wing area
$\kern4mm v\kern6mm$flight speed
$\kern4mm c_{D0}\kern2mm$zero-lift drag coefficient
$\kern4mm m\kern5mm$aircraft mass
$\kern4mm AR\kern1mm$wing aspect ratio
$\kern4mm \epsilon\kern6mm$Oswald factor
The figure below is the result plotted in R. Since I need to read the full matrix of values for the contour plot, the area of low speed and high bank angle is filled with the result of a strict penalty function, so please disregard the values to the right and below the red line.

Contour plot of height losses a A320 type aircraft in a 180° turn at sea level and MTOW (78 tons), no wind. X is bank angle in degrees and Y is flight speed in m/s. Own work.
As you can see, the minimum (approx. 170 m) is achieved right before stall at a high bank angle and speed. Unfortunately, you need the aerobatic version of the A320 to fly this safely.

Naive question. What constrains your acos numerator from becoming negative? Are those physically meaningless cases? Else your expression yields negatve bank angles?

@curious_cat: Negative bank angles show that the speed is not sufficient to create the required lift even in straight flight. The constraint is actually the speed, and this results in an upper bound for the lift coefficient.

At the stage where you define the term "smallest height loss for a given azimuth change", why cannot you cancel velocity & then that term becomes simply 1 / (R sin glide_angle) Since glide angle is known  you must minimize R to maximize  the term "smallest height loss for a given azimuth change". But R will be v^2/ (g tan bank_angle ). So to minimize R maximize  tan bank_angle & hence maximize bank_angle. So I might recommend bank as much as you can without stalling.  Am I making a blunder?

I have added another answer giving a detailed derivation of what I mean. Would love your critique on that if you could. Thanks!

@curious_cat: Steepest bank and AoA close to stall is the correct answer for the highest absolute turn rate. The smallest altitude loss is at minimum power loss, when $\frac{c_L^3}{c_D^2}$ is at its maximum. The best ratio of turn rate to altitude loss must be somewhere between both optima. Since speed is fixed, the end result depends on speed (and is certainly not always 45°).

The best bank angle for a gliding aircraft to use to optimize both turn rate and sink rate can be generalized as 45°.
The reason for this is that 45° is the point at which the vertical component of lift equals the horizontal component of lift.
In other words, a bank angle of 45° will produce the greatest centripetal turning force (horizontal lift) while maintaining the best sink rate (as a function of the vertical component of lift). A lesser bank angle will yield a better sink rate, but produce a lesser turn rate that decreases at a greater rate than the sink rate improves. Conversely, a greater bank angle will produce a better turn rate, but yield a greater sink rate that increases at a greater rate than the sink rate improves.
This phenomenon is purely a function of bank angle, wholly independent of other design or load factors, and therefore holds true for all fixed wing aircraft.
Edit: This may be a nominal answer, not accounting for minor variations in the L/D curves, but it satisfies my operational needs as a pilot experiencing an emergency where I will be maintaining "best glide" throughout a turn and my 45° bank is probably +/- 5°.

That's totally fine, answering your own question is actually encouraged

The highest turn rate regardless of sink speed can be obtained at maximum lift (i.e. right before stall), and the minimum altitude loss at the point when $\frac{c_L^3}{c_D^2}$ is at its maximum. The correct answer must be somewhere between both points and certainly depends on the fixed speed (which determines what lift reserves are available for turning). 45° will only be right at one specific speed.

There is no general solution, at least without further assumptions.
What I have up to now

The problem depends on both bank angle and airspeed

We can calculate the angular velocity (turn rate) from bank angle and airspeed

$$\omega = \frac{g \cdot tan~\theta}{v_H}$$

We can look up the vertical velocity from the (bank angle corrected) polar curve

$$v_V = \frac{f(v_H \cdot \sqrt{cos~\theta~~})}{\sqrt{cos~\theta~~}}$$

We can calculate the ratio $\frac{\omega}{v_V}$ that the asker wants to maximize for any airspeed and bank angle.

Without any knowledge of the polar in analytical form or assumptions we're stuck here.
Proof
Let $\theta$, $a_N$, $a_H$, $a_V$ be bank angle, normal, horizontal and vertical acceleration. Their relationships are:
$$a_V = a_N \cdot cos~\theta$$ $$a_H = a_N \cdot sin~\theta$$
with $a_V = 1g$
$$a_N = \frac{1}{cos~\theta} \cdot 1g$$
$$a_H = a_N \cdot sin~\theta = \frac{sin~\theta}{cos~\theta} \cdot 1g = g \cdot tan~\theta$$
Let $r$ and $\omega$ be turn radius and angular velocity. Let $v_H$ be the airspeed. The horizontal acceleration is the centripetal acceleration in our turn, so:
$$\frac{v_H^2}{r} = g \cdot tan~\theta$$
The angular velocity (turn rate) is:
$$\omega = \frac{v_H}{r} = \frac{g \cdot tan~\theta}{v_H}$$
Let $v_V$ be the sink rate which is a function $f$ of the airspeed. The function $f$ is usually given as polar curve. For load factors other than $1g$, we have to scale it by the square root of the load factor $k$.
$$k = \frac{1}{\sqrt{cos~\theta~~}}$$
$$v_V = k \cdot f \left( \frac{v_H}{k} \right)$$

Wouldn’t the mathematical problem look something like this?
$$V = L\cos\theta$$
$$H = L\sin\theta$$
$$R_S = k_1 (W - V)$$
$$R_T = k_2 H$$
Maximize $R_T / R_S$  with respect to $\theta$ given constant $L, k_1, k_2, W$
Notation:

$\theta$ is Bank Angle
$L$ is Lift
$H, V$ are Horizontal & Vertical Components of Lift
$W$ is Weight
$R_S$ is Sink Rate
$R_T$ is Turn Rate
$k_1, k_2$ are positive constants

The Math:
$R_T / R_S$ evaluates to
$$\frac{\frac{k_2 L}{k_1}\sin\theta}{W - L \sin\theta}$$
Maximise this expression with respect to $\theta$.
PS. When I do the math I get the $\theta$ that maximizes turn rate per sink rate as 90 degrees bank angle.
Obviously, I’m either messing up my math or my model. I must be wrong. Perhaps my blunder was to take lift $L$ as a constant? I suppose $L$ will change with bank too?
Also, I guess the stall characteristics should matter? Maybe that additional constraint would say bank at max angle that won’t stall you?
PS. This is just back of the envelope estimation. I’m probably being naive by not considering the complexities of the issue.

@JonathanWalters Could be. It might still disappear from the final expression.

Weird, I'm getting 90 degrees as the angle that maximizes the ration of turn rate to loss of altitude. I must be doing something wrong.

Lift is definitely not constant, because the vertical component must remain equal to weight, since there is no vertical acceleration throughout the manoeuvre.

You can't compute the sink rate from lift. Sink rate is equal to drag (times speed). (Induced) drag is needed in this equation.

Note that to assume drag is constant, you need to actually increase speed, because the best glide speed is actually best glide angle of attack and increases in the turn (by ~19% (⁴√2) for 45° bank).

Note that $k_1 (W - V)$ is a force (mass times acceleration), so if it is non-zero it represents the rate at which your sink rate accelerates. In the steady-state condition (constant sink rate) this expression has to be zero. -- Oh, wait, I think Jan Hudec already said this.

@DavidK Agreed. Too many problems in my derivation. Maybe I should just delete it.

Does it change something if you correct the sine at the denominator into cosine?

I'm adding another answer which looks way too simplistic but I cannot spot why it is wrong.
Here goes:
The term we want to maximize is "smallest height loss for a given azimuth change" & this can be shown to be the following borrowing @PeterKampfs logic:
$$ \frac{\omega}{v_z}=\frac{1}{R * sin \phi}$$
where $\phi$ is the glide angle.
Since glide angle is known you must minimize $R$ to maximize the term "smallest height loss for a given azimuth change". But $$R=\frac{v^2}{g*tan \theta}$$
So to minimize R maximize $tan \theta$ & hence maximize $\theta$. So I might recommend bank as much as you can.
But there's an additional constraint introduced by the fact that you must not stall (obviously). You must use maximum $\theta$ but not above $\theta_{stall}$
Let the un-banked stall speed be $v_{stall}$. Under a banked turn of $\theta$ the stall speed increases to:
$$v_{stall}^{banked}=v_{stall}*\sqrt{n}$$
where $n$ is the load factor.
$$n=\frac{L}{mg}=\frac{1}{cos \theta}$$
Therefore in a banked turn of $\theta$ the stall velocity will be:
$$v_{stall}^{banked}=\frac{v_{stall}}{\sqrt{cos \theta}}$$
Hence the max angle of bank would be the one you should use and that will be:
$$\theta=arccos \left( \left(\frac{v_{stall}}{v}\right)^2 \right) $$

@JonathanWalters Indeed. I'm using constant airspeed $v$. Where am I going wrong? Sorry if I'm not seeing the obvious.

@JonathanWalters No worries! But what do you feel about it? Can you spot any other flaws? If it is indeed correct I'm curious how it meshes with the other derivations given by Peter Kampf etc.,

@curious_cat: Since speed is fixed, $c_L$ varies with bank angle and we cannot assume that the glide angle is known. The airplane is somewhere on the polar, and not necessarily at the best $\frac{c_L}{c_D}$. Your result gives the highest possible turn rate regardless of altitude loss.


text:
Which US airport has the lowest circling MDH?
As I recently contemplated 14 CFR 135.223 and its implications, I began wondering about airports with exceptionally low circling minima.

For a little context and background, the part of §135.223 that sparked my imagination is the following exemption clause:

(1) The ceiling will be at least 1,500 feet above the lowest circling approach MDA

This got me thinking along these lines: "How often might one encounter both a circling MDH sufficiently low as to allow exemption from the fuel requirements of §135.223(a) and a ceiling high enough that the requirement of §91.169 to file an alternate airport remained in effect? Such a scenario would require a reported or forecast ceiling of 1900 ft and a circling MDH of 400 ft or less.
I know that there are at least a few airports with MDH of 400 ft or lower (which seems altogether too low). Thus, my question is:
Which airport in the United States has the lowest published circling Minimum Descent Height (MDH) for an instrument approach?
I initially wanted to apply this question equally to any airport worldwide. However, I have chosen to limit the scope of this question for two reasons.
Firstly, ICAO standards are more strict in this area than the TERPS standards that the US, Canada, and a few other parts of the world adhere to. Specifically, the TERPS standards for Cat A prescribe a minimum MDH of 350 ft, while ICAO prescribes a higher minimum of 394 ft.
Secondly, I recognize that discovering an ultimate answer to this question (i.e., "ABCD airport definitively has the lowest circling MDH") is difficult, especially if the question's scope included airports from multiple nations. For this reason, I have chosen to limit the scope of the question to the United States of America, where available databases should provide an opportunity for a definitive answer.

For example, the lowest MDH that I know of is Point Hope (PHO/PAPO) with a circling MDH of 361 ft for both the RNAV RWY 1 and RNAV RWY 19.

While this might be the lowest published MDH, I don't know whether or not it is.
I am looking for any answer that can factually establish which airport has the lowest published circling MDH, whether that might be Point Hope or some other airport.

If your only approach was a circling approach with a 1200' MDA, would 2000/3 get you out of needing an alternate, or in that case would you need an alternate with anything below a 2700' ceiling?

@RalphJ I.a.w. §91.169, given an IAP at the first airport of intended landing, and given a forecast of 2000/3 or greater, no alternate need be filed, regardless of IAP minima. I.a.w. §135.223, given a circling MDH of 1200, and given weather of 2000/3, sufficient fuel to fly to the alternate must be carried. That's what the regs say. I know, it doesn't make sense.

Part 135 operation is not part 91 operation. If you are a pilot in the USA you should know that basic difference.
A 135 operator would need 2700' ceiling to avoid planning an alternate if he CMDA is 1200.

Part of the problem is that most new approaches use an "assumed adverse obstacle" of 200feet above all terrain, because towers of 199ft do not need to be registered and there are many new cell towers erected each year.(also they don't need to survey every tree) The required obstacle clearance for circling is 300feet, plus the 200ft AAO = 500ft minimum unless some other factor comes in like land that is easy to visually survey for new towers and low potential for new trees, oil derricks or wind turbines in surrounding waters or land. Cat D and E min circling HAA is 550ft

@max §135 operations are still subject to §91 regulations in addition to the §135 regulations. §135 itself does not explicitly stated when an alternate need be filed.

J Walters Mostly incorrect. All IFR flights are required to file an alternate as default. Not planning an alt is considered an exception to the rule, and is only applied if the flight and operation can meet specific conditions, those conditions are different for 91 and 135. (135.223(b)) True that 135 does not explicitly state the alt must be filed before starting the flight or as part of the original flight plan, but it is implied by both the need to have it planned for fuel and because you need a clearance to the alt. 135/121 use op-specs, these can allow exceptions to 91.

@Max No, the need for an alternate is determined by both 91 and 135, as well as the operator's manual, and is not required by default but depending on forecasts. My company gives the conditions in which an alternate is required when I fly 135 flights, and these requirements are in accordance with 91 and 135 both. Do you have any experience operating under FAA 135 regulations?

Aa black and white fact, I have the regs in font of me, and plenty of specific education to properly read them, I also know how op-specs are developed and approved for 119 operators in general. The 135 reg overrides the 91 regs(just by being more restrictive) and an op-spec can override either, a POH has no bearing on this. As written an alt is always required as the default for IFR flight planning; not including an alt on an IFR plan is an exception only allowed for certain conditions. That you often have those qualifying exceptions does not mean that the exceptions become the default.

@Max I don't know where the POH comes in. By operator's manual I refer to the Air Carrier/Operator's manual as required by regulation (e.g., see §135.21). I believe you are arguing semantics (that an alternate is "always" required, except where not) and I am content to disagree with you and leave you to that on your own. The material point which I wished to emphasis is that §91 does apply as well to §135 operations. Our company's FAA accepted requirements for filing an alternate are derived from §91 for our §135 operations.

I took "operator's manual" as a pilots or aircraft operator's manual. I have always lumped the operator's manual you are talking about under the umbrella of "op-specs". My point is similar to yours in that there is a hierarchy to 135/121, 91, and op specs; I differ as not all 135 operations use your company's op specs and 135 covers alt. planning requirement. As for semantics, they literally determin the meaning of a statement, so yes in legal matters one is always arguing semantics. Which situation is the default matters when in a situation that has not been explicitly [or clearly] covered.

Ok, I see some of your confusion.The operator is required to have a manual distinct from the OpsSpecs that covers far more operational items. The Ops Specs are a defined list of operations for which an operator is or is not authorized. Each Ops Spec starts as a generic template with variances in detail for which the operator is authorized. The full scope of Ops Specs is available in the FSIMS. Re: semantics, the regulatory default is that you are not allowed to do anything, unless otherwise authorized. Talking about defaults is not helpful here where the situations have been clearly covered.

@MaxPower Another important distinction is that the Ops Specs are approved, while the manual is only accepted.

I'm going to partially answer this question. According to TERPS Section 2-7-1(b),

the minimum ROC in the circling approach OEA is 300 feet. Adjustments
must be applied as specified in paragraph 3-2-2.c.

OEA means Obstacle Evaluation Area, and ROC means Required Obstacle Clearance.
However, according to TERPS table 3-2-1 the minimum HAA (Height Above Airport) is 350 feet for CAT A aircraft.
If you look at the example given in TERPS Section 2-7 -1(c), it shows that the higher of the CMDA is published. Either 300 ft above the an obstacle located within the OEA, OR 350 ft above the airport. The 623 + 300 = 923 which rounds to 940. However, the 600 airport elevation + 350 = 950, but according to TERPS that rounds to 960. So 960 is published. The 960 is equal to a height of 360 ft AGL.
The way that I interpret this is that the lowest possible circling minimum descent altitude would be 360 ft above the airport assuming there are no obstacles in the OEA. However, the lowest circling height I've ever seen is around 440 AGL somewhere in New Mexico.  The obstacles are almost always going to dictate the minimum height unless there is an airport on a an elevation with no buildings or antennas nearby.
So, if you find an airport with a circling minimum height of 360 AGL, that is likely the airport with the lowest (or tied for the lowest) circling minimum in the US. The airport you found (Point Hope) at 361AGL is likely one of the lowest if not the lowest in the US.

This is a great way to answer the question on how low it can actually be.


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Is a physical lapel pin awarded for FAA Safety Wings?
I have recently got my PPL, and started the FAA Safety Wings program. After completion of each phase, are we awarded a physical lapel pin?

Yes, the FAA currently offers a lapel pin at the completion of each phase of the WINGS program.
This "reward" is not automatically awarded or delivered, the WINGS participant must request it. Upon completion of each of a phase, for example Basic, Advanced, or Master, the FAA Safety Team will include the following notice:

If you have not already done so, I encourage you to go to your My WINGS page at FAASafety.gov and claim your Team Member Reward -- a beautiful set of wings for your lapel.

The pins are sponsored and distributed by AVEMCO, in partnership with the FAA.
The pins are available in at least three finishes: bronze, silver, and gold, and a special sea-plane version is apparently available.

Source: own work

Beautiful!!! I just made the request and waiting for my lapel pin.... Thanks for the information.


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Where can fuel consumption data by aircraft type and route be found?
I am interested in building a data analytics platform for fuel consumption by aircraft type and route. I imagine this kind of information is readily available to airlines thanks to ACARS data. I need the data to build CO2 emissions models.
Are they any freely-available datasets or do I need to get this information from an airline directly?
Thanks!

Voting to close as this meets previously established criteria of "shopping" for a source of something.

I am providing an answer to my own question. However, hopefully, someone will be able to expand further on this.

For US Airlines, the bureau of transportation statistics is a reliable source. And the data is free!


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What are the stability characteristics of wide-cut jet fuels in storage?
Prompted by this comment.
Different hydrocarbon-based fuels have different stability characteristics when left to sit unused for long periods of time.  Avgas (and mogas too, for that matter) generally only stay good for about 6 months or so; in sharp contrast, kerosene-type (narrow-cut) jet fuels generally remain usable for around 2 years as-is, and about 10 years if treated with an antioxidant.
In very cold climates (places like Alaska, northern Canada, and [formerly] northern Russia), jets are often given wide-cut fuels, which trade the low vapor pressure (and resultant low high-temperature explosion hazard) of straight kerosenes for a much lower freezing point, as these sorts of places often get cold enough to freeze narrow-cut fuels, which is bad if it happens while the aircraft is in use.[citation needed]  In order to get this very-low freezing point, wide-cut fuels mix a lot of gasoline into the kerosene (Jet B, the predominant wide-cut fuel in North America, is an approximately-70:30 gasoline:kerosene mix).  As such, one might expect that the storage-stability characteristics of wide-cut jet fuels would be intermediate between those of straight gasolines and straight kerosenes.
What are the actual stability characteristics of wide-cut fuels in long-term storage?

Note: Russia, CIS, and China normally use TS1 fuel, which is still kerosene based, but with lighter fractions. Northern Europe runs Jet-A-1. Wide-cut Jet-B/JP-4 (almost the same) is mostly used in North America.

Avgas keeps a lot longer than autogas.  I would avoid using autogas more than 6 months old, but would have no problem flying with 2 year old avgas.

@JohnK: Interesting.  Does the TEL in avgas slow down the deterioration process?

I don't think it's the lead. It's just the base compound formulations are less volatile.  Modern auto gas keeps quite along time as well.  It takes years to get gum formation, and the octane level of car gas doesn't drop that much over time, a bit more than avgas. Ppl are still very conservative with it tho. It's so clean these days, got car gas all over my shirt a couple of years ago, and hours later my wife couldn't detect any smell.

Also, turbines may be way more tolerant of off fuel because they do not require more precise ratios like piston engines do.  Wider cuts may be more prone to absorbing water, but if it burns, a jet can use it.

The problem with fuels for turbine aircraft is that they can partially oxidize and form solid deposits, which then clog the fuel system.

Oxygenated mogas should be used in under 3 months, straight hydrocarbon fuel will last a year+, straight hydrocarbon with antioxidants(100ll) will last a couple of years. Depending on your cutoff and storage. Gums and varnish are caused by oxidation polymerization. Alcohols and ethers are added to mogas to increase oxidation (slightly more complete combustion, up to about 5 or 10% alcohol) but they kill the storage life so are kept separate at the refinery and blended when filling the delivery truck. Alcohol blends also have H2O issues that cause severe degradation in antiknock, +icing hazard

"alcohol free" mogas is still a "reformulated gasoline" by EPA standards it just substitutes an ether like MTBE for alcohol as the oxidant. It is slightly better than alcohol when it comes to moisture problems, but it still does not store like a straight hydrocarbon fuel. The stabilizer additives are very dilute antioxidant, but they are up against a 10% blend of oxygenator so it's kind of a lost cause, they might get you an extra month or two.


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Would an S-92 / CH-148 (or any helicopter) be able to land successfully after rolling over and eating someone with its main rotor in flight?
In the SCP Foundation story "Over/",1 the protagonist and his fiancée get kidnapped by some terrorists looking for information, who eventually throw the two of them out of a CH-148 (with parachutes, fortunately).  Immediately after the two exit, a mirror monster linked to the protagonist causes the copter to roll hard over; as the helicopter rolls towards inverted, one of the terrorists loses her grip and is finely minced by passage through the main rotor.  Protagonist and fiancée land safely, and the fiancée says that they'll need to run, because the remaining terrorists had managed to land the helicopter nearby.
I would not have expected a helicopter (CH-148 or otherwise) that'd taken an adult human through the main rotor while rolling through inverted to be capable of landing successfully, for a number of reasons:

Helicopter-main-rotor blades do not tend to react well to large, heavy objects passing through them, and chopping up an entire adult human would probably do some pretty nasty damage to the blades.
Helicopter main rotors are also fairly intolerant of mass imbalances resulting from blade damage; even a fairly-small section of one blade missing can result in an immediate loss of control of both the main rotor and the helicopter itself (often followed within seconds by the unbalanced main rotor tearing itself free of the helicopter altogether).
Finally, rolling rapidly through inverted would put all sorts of new and interesting forces on the main-rotor assembly in directions it was never designed to have force applied to it in (especially if, as seems to have been the case here, positive G was not maintained through the roll), potentially causing the main-rotor blades to strike the fuselage or the main-rotor mast to strike the adjacent structure2 (either of which would result in loss of control and main-rotor separation, not necessarily in that order) and certainly exacerbating the blade damage from an adult human going through the main rotor.

As just one example of this, in order for someone who loses her grip on the helicopter to pass through the main rotor, the main-rotor blades on that side of the helicopter would have to have a downward-pointing (in the chopper's frame of reference) lift vector (if the terrorist loses her grip while the main rotor is above her, then that side of the rotor has to accelerate downwards faster than the freefalling soon-to-be-mincee in order to catch up with her, necessitating that it be generating lift in the downwards direction; if she comes loose while the main rotor is under her, that side of the main rotor has to accelerate downwards slower than she does, requiring that it be generating lift in the upwards direction, which, as the main rotor being below its meal corresponds to the helicopter being inverted, requires that it be generating lift in the downwards direction from the helicopter's point of view).  This, like any other situation involving a helicopter's main rotor generating downwards-in-the-helicopter's-frame-of-reference lift, would cause the main-rotor blades on that side of the helicopter to cone downwards, towards the rest of the helicopter, and potentially collide with other helicopter structure.

Is it realistic that an S-92-family helicopter - or, for that matter, any helicopter - could (assuming no anomalous augmentations to the helicopter) roll hard through inverted, taking an adult human through the main rotor in the process, and still recover from the upset and make a safe emergency landing?

1: Not a typo.
2: While the S-92 family uses a fully-articulated main rotor, which is not normally susceptible to mast bumping, the extreme lateral forces that would be generated by a rotor disk tilted sufficiently to produce the roll rates involved in this situation might still damage the mast of even a fully-articulated main rotor, and mast bumping would still be an issue for many other types of helicopter in this sort of scenario.

All the human has to do is to grab on to the blade.  Then they will whirl around with it and not be minced.  ; )

A partial answer: some helicopter rotor blades are extremely fault tolerant. Just look at the grin on this lucky aviator and the blades on the chopper he has landed just a few a hours before:

More to read on the incident here: The Army Bestows This Rare Award to Pilots Who Crash Like Professionals
Alas, while the blades may survive the scenario, it is otherwise extremely unlikely, involving some very James Bond -like physics.

[reads article] HOLY CARP.

That's a Chinook, though, which has 2 lift rotors on it.  From the looks of it, only the rear one was damaged?  If that happened to a standard 1-rotor helicopter, it would be far less likely to be able to land safely.

@DarrelHoffman Chinook will fly just as poorly with one rotor as, say, a Blackhawk with no rotor.

Yeah, but the point once the damage occurs isn't to fly it, but just to land safely.  With one rotor still working, I'd think you'd still have at least a bit of lift to soften the landing.  With no rotor, you'd drop like a stone...

If we assume a single rotor would suffer same kind of damage as the featured Chinook, we can safely assume the controllability would be similar as in the Chinook's case. If we totally take away the only rotor, a helicopter will not fly at all. If we take away one of two rotors, the helicopter will not fly, at all, unless the rotors were coaxial, in which case the helicopter will fly very very poorly. In Chinook's case, losing one rotor will lead to such a great imbalance, the remaining rotor will be totally unusable.

No. I suspect having a human body come in to direct contact with the rotor disk 1) would not “finely mince” a person but instead cause severe blunt force trauma followed by flinging the body or body parts from the disk like a rag doll. 2) The impact would cause severe damage to one or more rotor blades, most likely causing separation of a blade from the base and gross imbalance of the rotor system.  This could eventually tear the aircraft apart in a manner similar to uncorrected ground resonance. 3) While (some) helicopters with fully articulated rotor systems are capable of limited inverted flight under positive loading, I don’t think an S-92 can do this.


text:
Why would changes in jet-engine power settings not result in changes in engine loudness?
My casual AAR reading has dredged up another counterintuitive detail of aircraft behaviour, this time regarding the relationship between engine power and engine loudness:

1.16.7.3 Sound Similar to Airplane Engines Increasing in Loudness
A review of the sound signatures associated with the rotating engines recorded by USAir flight 427’s CVR revealed that, throughout the 31-minute CVR recording, the engine’s sound signatures changed in frequency as the engine speed increased or decreased but remained constant in intensity (volume). During the upset sequence, the CAM channel of the accident CVR also recorded variations in the volume of the sound signatures associated with the engines. The sound spectrum study indicated that, at 1902:58.27 and 1903:02.3, the sounds on the CVR recording that were associated with both engines simultaneously increased in volume by about 30 percent; at the same time, the frequency of the engine sound signatures, and the volume of all other background noises, remained constant. The first increase signature occurred about 0.17 seconds after the first thump in the second set of thump sounds; the second increase occurred about 1 second later, just before the captain exclaimed “hang on.” Although FDR data indicated that the engine power settings changed during the accident sequence,201 the engine sounds (when identifiable) remained at the increased volume throughout the remainder of the CVR recording.

201 The FDR data indicated that, after a brief increase in engine power setting at 1903:02, the engine power settings began to decrease from the peak power setting recorded during the accident sequence (about 80 percent N1); by 1903:09, the engine power settings had decreased to about 30 percent N1, at which point they remained until ground impact.  [NTSB/AAR-99/01, page 132 (page 156 of the PDF).  Section heading bolded in original; all other emphasis mine.  Bolding emphasises that engine loudness generally remained constant despite changes in engine power; bold italics emphasise that the changes in loudness that did occur did not correspond to changes in engine power.1]

One would generally expect jet-engine loudness to correspond closely to engine power, for the following reasons (and quite possibly more as well):

At higher power settings, a much greater volume of air is forced through the engine and compressed and heated to a much greater degree than at low power settings, producing higher levels of engine airflow and increased tailpipe outlet pressure (observable in the cockpit as an increase in EPR, for engines so equipped), which one would expect to create more sound.
At higher power settings, the fan and compressor blades move faster and hit the incoming air harder, and the superheated gasses coming out of the combustion chamber move faster and hit the turbine blades harder, than at low power settings, which one would expect to create more sound.
At higher power settings, the core-exhaust jet is hotter and travelling faster than at low power settings, causing it to generate more vigorous turbulence for longer at its boundary with the bypass airflow, and the bypass air itself is hotter (due to undergoing a greater degree of compression from the faster-moving-and-harder-hitting fan blades) and faster-moving than at low power settings, causing it to generate more vigorous turbulence for longer at its boundary with the ambient air.  As turbulence at the interfaces of the differently-fast flow streams is a major source of noise (especially for engines not equipped with exhaust mixers), one would expect the increase in said turbulence at higher power settings to create more sound.
At higher power settings, more fuel is entering the combustion chamber and burning than at low power settings, which releases a greater amount of energy in so doing, which one would expect to create more sound.

Indeed, at very low engine speeds (for instance, when starting or shutting down an engine), a change in engine power does result in the expected change in loudness.  So why would the loudness of an engine in flight not depend on its power setting?

1: The anomalous increases in engine loudness were eventually deemed to reflect increased levels of fan noise resulting from alteration of engine-intake airflow due to the large sideslip angle attained by the aircraft during the accident sequence.


text:
Why is scud-running necessarily dangerous?
(Disclaimer: Please do not take the below as an endorsement of scud-running.)
Scud-running is when a pilot flying in poor weather operates their aircraft at a lower-than-normal height above terrain in order to remain clear of clouds and maintain visual contact with the ground.  This is generally considered a dangerous practice, having been linked to a number of fatal crashes.
What I’m curious about is why scud-running is necessarily so dangerous; by definition, it involves keeping one’s aircraft in clear air and maintaining visual reference with the ground, which should (in theory, at least) enable a pilot to maintain terrain clearance and avoid becoming disoriented (the latter of which is especially important for a non-instrument-rated pilot, for whom entry into cloud, unless followed by immediate exit from same, is not generally survivable).  Although flying closer to the ground does bring the aircraft closer to ground obstructions, such as trees, tall buildings, and cell-phone towers, these usually rise no more than 100 feet or so from ground level (a height low enough that one would be mere seconds from impacting the ground in any case), while the locations and heights of the rare exceptions are clearly marked on VFR charts, which should make it easy to give them a wide berth.
Even if the height between the ground and the clouds were to narrow to the point that an aircraft, to remain clear of cloud, would have to fly low enough to risk hitting low-height obstacles, this condition would be expected to become evident in plenty of time to turn away (a sudden step-down in the cloud base over a fairly-short distance is meteorologically unlikely, and would be visually obvious from a distance; a gradual lowering of the cloud base would still be obvious [from objects on the ground getting visibly closer and larger] long before one would risk striking low obstacles; and a terrain feature raising the ground into the clouds, such as a mountain, would be visible from a long way off).
Finally, one particular variant of scud-running is widely-used even in commercial passenger aviation: in a contact approach, a pilot gets ATC permission to discontinue a standard instrument approach and instead proceed at low altitude to the destination airport by visual reference to the ground, as long as the aircraft can stay out of the clouds and maintain at-least-1-mile visibility (equal to the bare VFR minima for uncontrolled airspace), and this is apparently safe enough for the FAA not to have banned contact approaches.
So why is scud-running necessarily a dangerous practice?

I think the reason scud-running is dangerous is because it has been "linked to a number of fatal crashes". If pilots weren't having accidents doing it, then it wouldn't be dangerous.

@GregHewgill: You're conflating cause and effect.

You are trying so hard to explain away the risk, but the very act of acknowledging and explaining all necessary risk mitigations demonstrates that you already have a clear understanding that there is an elevated danger level.  So what don’t you understand?

@MichaelHall: What I don't understand is why all the mitigations don't reduce the risk to a safe level.

And if followed to a T without any random negative events cropping up it can be done safely.  But the fact is that it is more dangerous than flying thousands of feet high in perfectly clear weather where the same risk mitigation’s are not necessary.  More things can go wrong and there is less time to react.  And the data supports this pretty obvious observation.

Have you ever heard the expression “flying isn’t inherently dangerous, it is just unforgiving of mistakes.”?

It's not necessarily dangerous.  I've flown hang gliders for hours and hours in ridge lift where it was often necessary to descend to lower altitude to avoid newly-forming orographic clouds and maintain visual contact with the ground.  Sounds like "scud running" to me.  And I'm still here!  Of course there was that one guy who accidentally lost contact with the ground and got turned around and almost landed in the ocean-- and that other guy who lost track of where he was and landed in that cemetery-- and that other guy who almost ran into that other guy when the cloud ceiling dropped suddenly-

"I've done it & I'm still here, ergo it can't be dangerous."  That way of thinking has let to more bad outcomes than most of us care to contemplate.  The fact that you got away with it doesn't mean that "it" isn't dangerous or unwise; that just means that it won't produce the bad outcome every time.  Don't confuse "got away with it" with "not dangerous."  EVER!!!

Does this answer your question? What are the difficulties and dangers faced by low flying aircraft? See also this question on inadvertent VMC into IMC flight, which is very dangerous.

Along a warm front, ceiling can change all the way down to the ground (fog).  Add in changes in ground level and towers (windmills lately), and it becomes unacceptably dangerous (unless you and your plane are ready for IFR, can monitor terrain, and know exactly where you are).  Very little difference between "clear" and "clould" in terms of humidity, and even haze can make visual navigation difficult.  Nothing like going up on days with 50-100 miles visibility if you fly recreational.

I once worked as a bush pilot on floats (my Most Fun Year), flying back and forth from a base in a large mining town to various lodges and fishing camps, ferrying wealthy Americans to their summer "camps" (cottages), and so on.  I spent many hours scud running (which I would define as flying under ceilings of less than 1000 ft and visibility of less than 3 mi).
The bush operator I was with was hard core; we used 50000:1 scale topographical maps to get around, where you might have to take 3 maps to cover a trip (aeronautical charts don't have the details you needed) and the flying in bad weather was straight pilotage, navigating from lake to lake, finger tracing the route on the map (It was 1990; GPS didn't exist, and Loran C was a thing and the Cessna 185 I was flying had it, but I didn't trust it. The only radio nav aid I ever used was the ADF, which I would tune to an AM radio station in town to listen to music and provide a bearing if necessary).
At the start of the season I would turn around and return on a trip when the visibility dropped below a couple miles.  By mid season I was flying in 1 mile and 500 ft, often cruising at 300 ft off the trees and a couple hundred feet below the cloud base to this place and that.
It is relatively safe in certain circumstances:

You know the operating area well, especially the hazards like towers and cable spans, etc.  My scud running trips were usually to places I'd been to many times and I got comfortable going in pretty bad conditions if I knew the route well. I would use higher limits if the trip was to somewhere new.
You have the ability to land anywhere if necessary.  Being on floats, with a land-able lake within 5-10 miles at any one time, provided ready escapes if things closed down.
It's much safer in the morning than the afternoon, because the dew point spread was increasing, not decreasing.  You avoided scud running in late afternoon whenever possible because whatever clag you were in was guaranteed to get worse as the temperature dropped.  When checking weather, the dew point spread was the most important bit of data.
When flying along river valleys, stay on the right side - other airplanes may be coming the other way, and you try to stay above the valley tops to avoid cable spans.  I would turn around if the cloud ceiling dropped to less than a couple hundred feet above the hill tops on each side of a valley and forced me down into the valley itself.

Now that is while bush flying on floats, over the Canadian Shield with a zillion lakes every which way.
Scud running in "civilization"?  Suicidal.  Towers everywhere.  200 ft towers, 500 ft towers. Towers are shown on charts, but new ones are popping up constantly.  If it's hilly country, you just stay away from the tallest hill tops, but in the flat lands they can be anywhere.
Power lines everywhere.  Clouds merging with hills.  Just too many things to run into.  Landing options few and far between (without damaging the plane at best) if the weather closes in.  I felt safe doing it on floats with lakes everywhere in a relatively remote area; I would never do it on wheels or in settled areas, for all the tea in China.
Yes contact approaches are allowed, but the exposure period is short, and they are still considered the most dangerous way to complete an IFR flight if the conditions are below normal VFR.

Likewise in mountains.  You're flying up a valley, have plenty of room between clouds & terrain when you start, but the ground keeps rising, and at some point you see there's no more room - and you don't have room to do a 180.